Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define Set(a, b) memset(a, b, sizeof (a))
#define For(i, j, k) for (int i = j; i <= k; ++ i)
#define Forr(i, j, k) for (int i = j; i >= k; -- i)

using namespace std;

typedef long long ll;

inline void File() {
	freopen("f.in", "r", stdin);
	freopen("f.out", "w", stdout);
}

const int N = 2.5e5 + 10;
int n; char s[N];
ll st[N], p[N];

const ll mod = 1000000000000000031ll;
const ll gkk = 1000000000000000003ll;
const ll hany01 = 821428571428571454ll;

ll mul(ll a, ll b) { return (__int128) a * b % mod; }
ll add(ll a, ll b) { return (a += b) >= mod ? a - mod : a; }

unordered_map<ll, int> mp;

int main() {

	cin >> n, scanf("%s", s + 1);

	p[0] = 1;
	For(i, 1, n) {
		st[i] = st[i - 1], p[i] = p[i - 1];
		if (s[i] == '+') st[i] = add(st[i], p[i]);
		if (s[i] == '-') st[i] = add(st[i], mod - p[i]);
		if (s[i] == '>') p[i] = mul(p[i], gkk);
		if (s[i] == '<') p[i] = mul(p[i], hany01);
	}

	ll ans = 0;
	Forr(i, n, 1) {
		++ mp[st[i]];
		ans += mp[add(mul(st[n], p[i - 1]), st[i - 1])];
	}

	cout << ans << endl;
	return 0;
}